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Aleonysh [2.5K]
3 years ago
8

Please please please help

Mathematics
1 answer:
Neko [114]3 years ago
6 0
The triangle in the front has base 2+2 and height 4.5 so area \frac 1 2(2+2)(4.5)

The top fin is \frac 1 2(2.25+0.75)(3)

The back fin is \frac 1 2(4.5)(3+5)

The bottom fin \frac 1 2 (1+.5)(1)

The back of the fish by the tail, \frac 1 2 (2+2)(4.5)

The four little triangles, 4 (\frac 1 2 )(.5)(1.5)

Two thin rectangles, 2(.5)(5)

One big rectangle  (2+2)(1.5+5+1.5)


Adding it all up

\frac 1 2 ( (2+2)(4.5) + (2.25+0.75)(3)+ (4.5)(3+5)+(1+.5)(1)+(2+2)(4.5)+4(.5)(1.5)) +2(.5)(5)+(2+2)(1.5+5+1.5) = 122.5

Probably a mistake in there somewhere; check it good.


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Answer:

6√5

Step-by-step explanation:

√15 x √12

=√3x √5 x 2 x √3

=3 x 2 x √5

=6 x √5

=6√5

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4 years ago
Please answer the sum correctly
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33tomato and 23 chicken that equals 56 and hasn’t a difference of 10. Hope that helps

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I need the right answer please
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.. {-9, -6, 4, 7}

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The 4th selection {4, 7} is the best of those offered, but it is not correct.

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3 years ago
What is the fraction for 7.71
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4 0
3 years ago
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Question 7 of 10
andriy [413]

Answer:

C. n = 90; p = 0.8

Step-by-step explanation:

According to the Central Limit Theorem, the distribution of the sample means will be approximately normally distributed when the sample size, 'n', is equal to or larger than 30, and the shape of sample distribution of sample proportions with a population proportion, 'p' is normal IF n·p ≥ 10 and n·(1 - p) ≥ 10

Analyzing  the given options, we have;

A. n = 45, p = 0.8

∴ n·p = 45 × 0.8 = 36 > 10

n·(1 - p) = 45 × (1 - 0.8) = 9 < 10

Given that for n = 45, p = 0.8, n·(1 - p) = 9 < 10, a normal distribution can not be used to approximate the sampling distribution

B. n = 90, p = 0.9

∴ n·p = 90 × 0.9 = 81 > 10

n·(1 - p) = 90 × (1 - 0.9) = 9 < 10

Given that for n = 90, p = 0.9, n·(1 - p) = 9  < 10, a normal distribution can not be used to approximate the sampling distribution

C. n = 90, p = 0.8

∴ n·p = 90 × 0.8 = 72 > 10

n·(1 - p) = 90 × (1 - 0.8) = 18 > 10

Given that for n = 90, p = 0.9, n·(1 - p) = 18 > 10, a normal distribution can be used to approximate the sampling distribution

D. n = 45, p = 0.9

∴ n·p = 45 × 0.9 = 40.5 > 10

n·(1 - p) = 45 × (1 - 0.9) = 4.5 < 10

Given that for n = 45, p = 0.9, n·(1 - p) = 4.5 < 10, a normal distribution can not be used to approximate the sampling distribution

A sampling distribution Normal Curve

45 × (1 - 0.8) = 9

90 × (1 - 0.9) = 9

90 × (1 - 0.8) = 18

45 × (1 - 0.9) = 4.5

Now we will investigate the shape of the sampling distribution of sample means. When we were discussing the sampling distribution of sample proportions, we said that this distribution is approximately normal if np ≥ 10 and n(1 – p) ≥ 10. In other words

Therefore;

A normal curve can be used to approximate the sampling distribution of only option C. n = 90; p = 0.8

3 0
3 years ago
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